解决方案很简单，只需使用CascadeType。MERGE而不是CascadeType。PERSIST或CascadeType.ALL。

我也遇到过同样的问题和CascadeType。MERGE对我很有效。

我希望你已经整理好了。

也许这是OpenJPA的bug，当回滚时它重置了@Version字段，但pcVersionInit保持true。我有一个声明@Version字段的AbstraceEntity。我可以通过重置pcVersionInit字段来解决它。但这不是一个好主意。我认为它不工作时，级联坚持实体。

    private static Field PC_VERSION_INIT = null;
    static {
        try {
            PC_VERSION_INIT = AbstractEntity.class.getDeclaredField("pcVersionInit");
            PC_VERSION_INIT.setAccessible(true);
        } catch (NoSuchFieldException | SecurityException e) {
        }
    }

    public T call(final EntityManager em) {
                if (PC_VERSION_INIT != null && isDetached(entity)) {
                    try {
                        PC_VERSION_INIT.set(entity, false);
                    } catch (IllegalArgumentException | IllegalAccessException e) {
                    }
                }
                em.persist(entity);
                return entity;
            }

            /**
             * @param entity
             * @param detached
             * @return
             */
            private boolean isDetached(final Object entity) {
                if (entity instanceof PersistenceCapable) {
                    PersistenceCapable pc = (PersistenceCapable) entity;
                    if (pc.pcIsDetached() == Boolean.TRUE) {
                        return true;
                    }
                }
                return false;
            }

如果没有任何帮助，并且您仍然得到这个异常，请检查equals()方法—并且不要在其中包含子集合。特别是当你有嵌入式集合的深层结构时(例如A包含B, B包含c，等等)。

以Account ->为例:

  public class Account {

    private Long id;
    private String accountName;
    private Set<Transaction> transactions;

    @Override
    public boolean equals(Object obj) {
      if (this == obj)
        return true;
      if (obj == null)
        return false;
      if (!(obj instanceof Account))
        return false;
      Account other = (Account) obj;
      return Objects.equals(this.id, other.id)
          && Objects.equals(this.accountName, other.accountName)
          && Objects.equals(this.transactions, other.transactions); // <--- REMOVE THIS!
    }
  }

在上面的例子中，从equals()检查中删除事务。这是因为hibernate将暗示您不尝试更新旧对象，而是在更改子集合上的元素时传递一个新对象来持久化。 当然，这种解决方案并不适用于所有应用程序，您应该仔细设计想要包含在equals和hashCode方法中的内容。

使用合并是有风险和棘手的，所以在您的情况下，这是一种肮脏的变通方法。您至少需要记住，当您将一个实体对象传递给merge时，它将停止附加到事务，而是返回一个新的、现在已附加的实体。这意味着如果任何人仍然拥有旧的实体对象，那么对它的更改将被无声地忽略并在提交时丢弃。

You are not showing the complete code here, so I cannot double-check your transaction pattern. One way to get to a situation like this is if you don't have a transaction active when executing the merge and persist. In that case persistence provider is expected to open a new transaction for every JPA operation you perform and immediately commit and close it before the call returns. If this is the case, the merge would be run in a first transaction and then after the merge method returns, the transaction is completed and closed and the returned entity is now detached. The persist below it would then open a second transaction, and trying to refer to an entity that is detached, giving an exception. Always wrap your code inside a transaction unless you know very well what you are doing.

使用容器管理的事务，它看起来像这样。注意:这假设方法在会话bean中，并通过本地或远程接口调用。

@TransactionAttribute(TransactionAttributeType.REQUIRED)
public void storeAccount(Account account) {
    ...

    if (account.getId()!=null) {
        account = entityManager.merge(account);
    }

    Transaction transaction = new Transaction(account,"other stuff");

    entityManager.persist(account);
}

cascadeType.MERGE,fetch= FetchType.LAZY

@OneToMany(mappedBy = "xxxx", cascade={CascadeType.MERGE, CascadeType.PERSIST, CascadeType.REMOVE})

为我工作。